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E E 2315 Circuits I
Lecture 2 - Circuit Elements
and Essential Laws
Five Fundamental Elements
• Ideal Voltage Sources
– Independent
– Dependent
• Ideal Current Sources
– Independent
– Dependent
• Resistors
• Inductors (to be introduced later)
• Capacitors (to be introduced later)
Independent Voltage Source
• Voltage may be
constant or
time-dependent
• Delivers
nominal
terminal
voltage under
all conditions
Positive Terminal
Vg
Negative Terminal
Independent Current Source
• Current may be
constant or timedependent
• Delivers nominal
terminal current
under all
conditions
Negative Node
Ig
Positive Node
Voltage-Controlled Dependent Voltage
Source
• Terminal voltage
is a function of the
voltage drop of a
different branch
• Delivers nominal
terminal voltage
under all
conditions
Positive Terminal
v
Negative Terminal
+
v
-
Current-Controlled Dependent
Voltage Source
• Terminal voltage is
a function of the
current flow in a
different branch
• Delivers nominal
terminal voltage
under all
conditions
Positive Terminal
i
i
Negative Terminal
Voltage-Controlled Dependent
Current Source
• Current is a
function of the
voltage drop of a
different branch
• Delivers nominal
terminal current
under all
conditions
Negative Node
v
Positive Node
+
v
-
Current-Controlled Dependent
Current Source
• Source current is a
function of the
current flow in a
different branch
• Delivers nominal
terminal current
under all
conditions
Negative Node
i
Positive Node
i
Electrical Resistance
(Ohm’s Law)
• Electrical resistance is
the ratio of voltage
drop across a resistor to
current flow through
the resistor.
• Polarities are governed
by the passive sign
convention.
R
i
+
v
-
v
R
i
Power Consumed by Resistors
• Resistors
consume
power.
• v and i are both
positive or
both negative.
R
i
+
v
-
p  v i
v  R i
v
i
R
Conductance Defined
• Conductance is the
reciprocal of
resistance.
• The units of
conductance are
called siemens (S)
• The circuit symbol
is G
1
G
R
i  v G
v i
G
Creating a Circuit Model
• A circuit model is usually two or more
circuit elements that are connected.
• A circuit model may have active
elements (sources) as well as passive
elements (such as resistors).
• By the assumption that electric signal
propagation is instantaneous in a
circuit, our circuit model has lumped
parameters.
Example of a Circuit Model
1000 ft AWG 14
Copper Wire
100 W
Lamp
120 V Battery
0.25 
2.57 
144 
120 V
2.57 
Kirchhoff’s Voltage Law
• The sum of the voltage drops around a
closed path is zero.
• Example: -120 + V1 + V2 + V3 + V4 = 0
0.25 
+ V1 -
2.57 
+ V2 -
120 V
2.57 
- V4 +
+
V3
-
144 
Kirchhoff’s Current Law
• A node is a point where two or more
circuit elements are connected together.
• The sum of the currents leaving a node
is zero.
I1
I4
I3
I2
Apply KCL to Example
I1
Is
0.25  I1
+ V1 -
I2
2.57  I2
120 V
2.57 
Is
I4
I3
+ V2 +
V3
-
- V4 + I4
144 
I3
Combine KVL, KCL & Ohm’s Law
I1
Is
0.25  I1
+ V1 -
I2
2.57  I2
120 V
2.57 
Is
I4
I3
+ V2 +
V3
-
144 
I3
- V4 + I4
120  0.25  I s  2.57  I s  144  I s  2.57  I s
Lamp Voltage & Battery Voltage
I1
Is
0.25  I1
+ V1 -
I2
+
2.57  I2
Vb
120 V
-
Is
I4
I3
+ V2 -
2.57 
+
V3
-
144 
I3
- V4 + I4
V3  144  I s  115.67V
Vb  (2.57  2  144)  0.803  119.8V
Battery Power and Lamp Power
1000 ft AWG 14
Copper Wire
100 W
Lamp
120 V Battery
Pb  119.8V  0.8033 A  96.23W
Loss:
Efficiency:
Ploss  Pb  Pl  3.32W
Using Loops to Write Equations
vb
R2
+ v2 va
a
+
v1
-
R1
b
+
v3
-
R3
c
KVL @Loop a:
KVL @ Loop b:
KVL @ Loop c:
va  v2  v1  0
va  v2  vb  v3  0
Loop c equation same as a & b combined.
Using Nodes to Write Equations
y
ia
va
i2
R2
i2
x
+ v2 -
ib
vb
z
i3
i1
+
v1
-
+
v3
-
R1
i1
ia
ib
R3
i3
w
KCL @ Node x:
KCL @ Node y:
KCL @ Node z:
KCL @ Node w:
i2  i1  ib  0
ia  i2  0
ia  i1  i3  0 <== Redundant
Combining the Equations
• There are 5 circuit elements in the
problem.
• va and vb are known.
• R1, R2 and R3 are known.
• v1, v2 and v3 are unknowns.
• ia, ib, i1, i2 and i3 are unknowns.
• There are 2 loop (KVL) equations.
• There are 3 node (KCL) equations.
• There are 3 Ohm’s Law equations.
• There are 8 unknowns and 8 equations.
Working with Dependent
Sources
4
i
48 V
KVL @ left loop:
i
+
vo
-
3
i
48V  4  i  3  io
KCL @ top right node:
Substitute and solve:
i  3 A vo  36V
Example 1 (1/3)
20 A
Ie
50 V
By KCL:
30 A
Ic
+ Va 25 
Id
+ + Vb - +
Vd 10 
Vc 50 V If
ie  20 A, id  30 A, i f  30 A, ic  10 A
By Ohm’s Law:
Example 1
(2/3)
Ie
50 V
By KVL:
20 A
30 A
Ic
+ Va 25 
Id
+ + Vb - +
Vd 10 
Vc 50 V If
Va  300V , Vb  600V
Power:
Pb  600V  30 A  18.0 kW
Example 1
(3/3)
Ie
50 V
20 A
30 A
Ic
+ Va 25 
Id
+ + Vb - +
Vd 10 
Vc 50 V If
Pc  250V   10 A   2.5 kW
Pd  300V  30 A  9.0 kW
Pf  50V  30 A  1.5 kW
Example 2 (1/4)
1
3A
I
12 
84 V
R
4
8
8
12 
Find Source Current, I, and Resistance, R.
Example 2 (2/4)
1
3A
I
84 V
12 
6A
8
Ohm’s Law: 36 V
+
36 V
+4
48 V
12 
R
8
KVL: 48 V Ohm’s Law: 6 A
Example 2
(3/4)
1
3A
I
84 V
12 
6A
8
KCL: 3 A
Ohm’s Law: 12 V
+
36 V
3 A -12 V+
+4 +
48 V 60 V
- 12 
R
8
KVL: 60 V
Example 2 (4/4)
1
3A
I
84 V
Ohm’s Law: 3 A
KVL: 24 V
12 
6A
8
++
36 V 24 V
3 A -12 V+
+4 +
48 V 60 V
- 12 
Ohm’s Law: R=3 
6A
R
3A
8
KCL: I=9 A